# Showing a function is not differentiable at $(0,0)$

Let $\displaystyle f(x,y)=\begin{cases} \frac{x^3+y^4}{x^2+y^2} \text{ if } (x,y) \neq (0,0)\\ 0 \text{ if } (x,y)=(0,0). \end{cases}$ Show this is not differentiable at $(0,0)$.

My strategy is to compare $D_uf(p)$ and $\nabla f(p)$ and hopefully $D_uf(p)\neq\nabla f(p)$.

So I computed the directional derivative:
Let $\displaystyle u=(u_1, u_2)$. Then $\displaystyle D_uf(p)=\lim\limits_{t \to 0}\frac{f((0,0)+t(u_1,u_2))-f((0,0))}{t}=u_1^3$

"Work" $\displaystyle\frac{t^3u_1^3+t^4u_2^4}{t(t^2u_1^2+t^2u_2^2)}=\frac{u_1^3+tu_2}{u_1^2+u_2^2}=u_1^3+tu_2$

My problem is computing the gradient. I worked it out by hand and got some stuff that looked like I would end up with $\frac{0}{0}$. Is there another way to compute the gradient that I am missing.

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We want to show $f$ is not differentiable. We suppose to the contrary. Then $\nabla f$ exists, and $\nabla f\cdot u=D_u f$, for all $u$. That is, the tangent plane is well defined. If $\nabla f$ exists, than you have a steepest direction and a reverse, and all the other directions lead to various inclinations as if you were on a plane. But if $f$ is not differentiable, this is not the case. $D_uf$ may exist for all $u$, but it won't be predicted like a plane.

If you graph it, like here: http://www.wolframalpha.com/input/?i=%28x%5E3%2By%5E4%29%2F%28x%5E2%2By%5E2%29+

then you see the plane is not well defined at the origin. There's too many slopes coming in.

Let's continue with our proof by contradiction. If $\nabla f$ exists, then we have $$$u_1^3=D_uf(0,0)=\nabla f(0,0)\cdot u=f_x u_1+f_y u_2.$$ But this is impossible for constants$f_x(0,0)$,$f_y(0,0)$. - Why is it impossible for those constants? – abet Mar 31 at 21:01 add comment Let's use the direct definition of differentiability - that is, existence of a matrix,$\mathbf{J(x_a)}$, such that $$\lim_{\mathbf{h\to 0}} \frac{\mathbf{f(x_a+h)-f(x_a)-J(x_a)h}}{||\mathbf{h}||} = \mathbf{0}$$ In our case, the matrix is actually a vector. Using radial coordinates, this is easier. We have$x=r\cos\theta$and$y=r\sin\theta$, and so$x^2+y^2=r^2$. So $$f(r,\theta) = r\cos^3\theta+r^2\sin^4\theta$$ Which can be confirmed to be accurate$\forall (r,\theta)$. Now, at$\mathbf{x_a=0}$, we have$r=0$, and$\mathbf{h=r}=(x,y)=(r\cos\theta,r\sin\theta)$. So we ask if there exists a matrix$\mathbf{J}$such that $$\lim_{r\to 0} \frac{r\cos^3\theta+r^2\sin^4\theta-\mathbf{J(0)\cdot r}}{r} = 0$$ Note that$\mathbf{J(0)}$cannot depend on$r$or$\theta$. If we let$\mathbf{J(0)}=(a,b), then \begin{align} \lim_{r\to 0} \frac{r\cos^3\theta+r^2\sin^4\theta-ar\cos\theta-br\sin\theta}{r} &=\lim_{r\to 0} \cos^3\theta+r\sin^4\theta-a\cos\theta-b\sin\theta\\ &=\cos^3\theta-a\cos\theta-b\sin\theta \end{align} From here, it is easy to see that no such values ofa$and$b\$ may exist such that the limit equals zero, and so the function is not differentiable.

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As well, you can calculate the Jacobian matrix. Its transpose is the gradient. You have $$\vec{\nabla}f(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \end{bmatrix}^T.$$